5.12.6: Chapter Test
Perform the indicated operation for the expression: \((3{x^2} + 2x + 1) - ({x^2} - 2 + 2) + x(3x + 11)\)
Solve the linear equations using properties of equations: \(x(x - 2) + = 9x + x(x - 6)\)
It costs 55 cents for a stamp. Construct a linear equation and solve how much it cost to buy 50 stamps.
Solve the formula \(A = \frac
Callstack:
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Solve the inequality \(-2x < -6\), graph the solution on the number line, and write the solution in interval notation.
Construct a linear inequality to solve the application: Bella wants to buy a round of shakes for her friends. It will cost $4.75 per shake, including tip and tax. Her budget is $50. What is the maximum number of friends Bella can buy shakes for?
Manneken Pis is a famous statue in Brussels, Belgium. It is 24 inches tall and weighs 37.5 pounds. The average man is 69 inches tall and weighs 198 pounds. Is Manneken Pis proportional to the average male?
Graph \(y = \frac{1}{4}x + 1\) by plotting points.
Graph the linear inequality: \(y \ge 2x + 7\)
Graph and list the solutions to the quadratic equation \({x^2} - 36 = 0\).
Solve \({x^2} - 49 = 0\) by factoring.
Solve \({x^2} + 5x + 2 = 0\) using the quadratic formula.
Evaluate the function \(f(x) = -3x + 21\) at the values \(f (-2)\), \(f (-1)\), \(f (0)\), \(f(1)\), and \(f (2)\).
Use the vertical line test to determine if the given graph represents a function.
Use the graph shown to find the domain and the range.
Graph \(2y = x - 4\) using the intercepts.
Use the slope formula to find the slope of the line between (1, 4) and (3, 5).
Identify the slope and \(y\)-intercept of \(y - \frac{2}{3}x = 1\).
Graph the line of \(y = \frac{2}{3}x + 1\) using its slope and \(y\)-intercept.
The equation \(C = 4.50 + 15p\), models the cost of visiting the Cat Café in San Diego for one hour. \(C\), in dollars, is the total cost and the cost per person, \(p\), is $15 plus a $4.50 reservation fee.
Find the payment for two people.
Find the payment for five people.
Interpret the slope and \(C\)-intercept of the equation.
Graph the equation.
Solve the system of equations by graphing.
\(\left\{ {\begin{array}{*{20}{l}}{y - 3x = 3}\\{y = - 6x + 12}\end{array}} \right.\)
Solve the system of equations by substitution.
\(\left\{ {\begin{array}{*{20}{l}}{y = 5x - 5}\\{ - 3x + y = - 3}\end{array}} \right.\)
Solve the systems of equations by elimination.
\(\left\{ {\begin{array}{*{20}{l}}{y = \frac{1}{3}x - 6}\\{y = x - 3}\end{array}} \right.\)
Anna goes to the concession stand at a movie theater.  During intermission, Isabelle goes to the concession stand.  What is the cost of one popcorn, and the cost of one large soda?
Solve the systems of linear equations by graphing.
\(\left\{ {\begin{array}{*{20}{l}}{y > \frac{1}{3}x - 1}\\{y < x - 3}\end{array}} \right.\)
Juliette is selling fresh lemonade and cupcakes. She sells a cup of lemonade for $2 and a cupcake for $3. She needs to make at least $100 to donate to the local cat sanctuary. She needs to sell at least 20 cups of lemonade.
Write a system of inequalities to model this situation.
Graph the system.
Could she sell 30 cups of lemonade and 10 cupcakes and make $100?
Could she sell 20 cups of lemonade and 30 cupcakes and make $100?
A toy maker makes exactly two toys out of wood; the Box (\(x\)) and the Bat (\(y\)). He makes $5 per Box and $6 per Bat. Each Box requires 30 ounces of wood, and each Bat requires 45 ounces of wood. Today the toy maker has 270 ounces of wood available. The toy maker also only makes 8 wooden toys per day. To maximize profit, how many of each wooden toy should the toy maker make?
Find the objective function.
Write the constraints as a system of inequalities.
Graph of the system of inequalities.
Find the value of the objective function at each corner point of the graphed region.
To maximize profit, how many of each toy should the toymaker make?


33.1: Meaning of Exponents
Lesson
Let's see how exponents show repeated multiplication.
Exercise \(\PageIndex{1}\): Notice and Wonder: Dots and Lines
What do you notice? What do you wonder?
Exercise \(\PageIndex{2}\): The Genie's Offer
You find a brass bottle that looks really old. When you rub some dirt off of the bottle, a genie appears! The genie offers you a reward. You must choose one:
- $50,000; or
- A magical $1 coin. The coin will turn into two coins on the first day. The two coins will turn into four coins on the second day. The four coins will double to 8 coins on the third day. The genie explains the doubling will continue for 28 days.
- The number of coins on the third day will be \(2\cdot 2\cdot 2\). Write an equivalent expression using exponents.
- What do \(2^{5}\) and \(2^{6}\) represent in this situation? Evaluate \(2^{5}\) and \(2^{6}\) without a calculator.
- How many days would it take for the number of magical coins to exceed $50,000?
- Will the value of the magical coins exceed a million dollars within the 28 days? Explain or show your reasoning.
Explore the applet. (Why do you think it stops?)
Are you ready for more?
A scientist is growing a colony of bacteria in a petri dish. She knows that the bacteria are growing and that the number of bacteria doubles every hour.
When she leaves the lab at 5 p.m., there are 100 bacteria in the dish. When she comes back the next morning at 9 a.m., the dish is completely full of bacteria. At what time was the dish half full?
Exercise \(\PageIndex{3}\): Make 81
- Here are some expressions. All but one of them equals 16. Find the one that is not equal to 16 and explain how you know.
\(2^{3}\cdot 2\qquad 4^{2}\qquad\frac{2^{5}}{2}\qquad 8^{2}\)
- Write three expressions containing exponents so that each expression equals 81.
Summary
When we write an expression like \(2^{n}\), we call \(n\) the exponent.
If \(n\) is a positive whole number, it tells how many factors of 2 we should multiply to find the value of the expression. For example, \(2^{1}=2\), and \(2^{5}=2\cdot 2\cdot 2\cdot 2\cdot 2\).
There are different ways to say \(2^{5}\). We can say “two raised to the power of five” or “two to the fifth power” or just “two to the fifth.”
Practice
Exercise \(\PageIndex{4}\)
Select all the expressions that are equivalent to \(64\).
- \(2^{6}\)
- \(2^{8}\)
- \(4^{3}\)
- \(8^{2}\)
- \(16^{4}\)
- \(32^{2}\)
Exercise \(\PageIndex{5}\)
Select all the expressions that equal \(3^{4}\).
- \(7\)
- \(4^{3}\)
- \(12\)
- \(81\)
- \(64\)
- \(9^{2}\)
Exercise \(\PageIndex{6}\)
\(4^{5}\) is equal to 1,024. Evaluate each expression.
- \(4^{6}\)
- \(4^{4}\)
- \(4^{3}\cdot 4^{2}\)
Exercise \(\PageIndex{7}\)
\(6^{3}=216\). Using exponents, write three more expressions whose value is \(216\).
Exercise \(\PageIndex{8}\)
Find two different ways to rewrite \(3xy+6yz\) using the distributive property.
(From Unit 6.2.6)
Exercise \(\PageIndex{9}\)
Solve each equation.
\(a-2.01=5.5\)
\(b+2.01=5.5\)
\(10c=13.71\)
\(100d=13.71\)
(From Unit 6.1.5)
Exercise \(\PageIndex{10}\)
Which expressions represent the total area of the large rectangle? Select all that apply.
- \(6(m+n)\)
- \(6n+m\)
- \(6n+6m\)
- \(6mn\)
- \((n+m)6\)
(From Unit 6.2.5)
Exercise \(\PageIndex{11}\)
Is each statement true or false? Explain your reasoning.
- \(\frac{45}{100}\cdot 72=\frac{45}{72}\cdot 100\)
- \(16\)% of \(250\) is equal to \(250\)% of \(16\)
(From Unit 3.4.7)


6.2: Logarithmic Scales in Natural Sciences
In 2010, a major earthquake struck Haiti, destroying or damaging over 285,000 homes [1] . One year later, another, stronger earthquake devastated Honshu, Japan, destroying or damaging over 332,000 buildings, [2] like those shown in the image above. Even though both caused substantial damage, the earthquake in Japan was 100 times as strong as the earthquake in Haiti. How do we know? The magnitudes of earthquakes are measured on a scale known as the Richter Scale. The Haitian earthquake registered a 7.0 on the Richter Scale [3] whereas the Japanese earthquake registered a 9.0. [4]
The Richter Scale is a base-ten logarithmic scale. In other words, an earthquake of magnitude 8 is not twice as great as an earthquake of magnitude 4. It is 108-4=104=10,000 times as great! In this section, we will investigate the nature of the Richter Scale and the base-ten function upon which it depends.
Converting from Logarithmic to Exponential Form
In order to analyze the magnitude of earthquakes or compare the magnitudes of two different earthquakes, we need to be able to convert between logarithmic and exponential form. For example, suppose the amount of energy released from one earthquake were 500 times greater than the amount of energy released from another. We want to calculate the difference in magnitude. The equation that represents this problem is \(10^{x}=500\), where \(x\) the difference in magnitudes on the Richter Scale. How would we solve for \(x\)?
We have not yet learned a method for solving exponential equations. None of the algebraic tools discussed so far is sufficient to solve \(10^{x}=500\). We know that \(10^{2}=100\) and \(10^{3}=1000\), so it is clear that \(x\) must be some value between 2 and 3, since \(y=10^{x}\) increasing. We can examine a graph, below , to better estimate the solution.
Using Common Logarithms
Sometimes we may see a logarithm written without a base. In this case, we assume that the base is 10. In other words, the expression \(\log (x)\) means \(\log _{10}(x)\). We call a base-10 logarithm a common logarithm . Common logarithms are used to measure the Richter Scale mentioned at the beginning of the section. Scales for measuring the brightness of stars and the pH of acids and bases also use common logarithms.
Definition: Common Logarithm
A common logarithm is a logarithm with base 10. We write \(\log _{10}(x)\) simply as \(\log (x)\). The common logarithm of a positive number \(x\) satisfies the following definition.
For \(x>0\)
\(y=\log (x)\) is equivalent to \(10^{y}=x\)
We read \(\log (x)\) as, “the logarithm with base 10 of \(x\)” or “log base 10 of \(x\).”
The logarithm \(y\) is the exponent to which 10 must be raised to get \(x\).
How to
Given a common logarithm of the form \(y=\log (x)\), evaluate it mentally.
- Rewrite the argument x as a power of \(10: 10^{y}=x\).
- Use previous knowledge of powers of 10 to identify \(y\) by asking, “To what exponent must 10 be raised in order to get \(y\)?
Example \(\PageIndex{1}\): Finding value of a common Logarithm Mentally
Evaluate \(y=\log (1000)\) without using a calculator.
Solution
First we rewrite the logarithm in exponential form: \(10^{y}=1000\). Next, we ask, “To what exponent must 10 be raised in order to get 1000?” We know
\[\begin{aligned}
10^{3}&=1000\\
\therefore \log (1000)&=3
\end{aligned} \nonumber \]
Try it Now 4
Evaluate \(y=\log (1,000,000)\).
How to
Given a common logarithm with the form \(y=\log (x)\), evaluate it using a calculator.
- Press [LOG].
- Enter the value given for x followed by [ ) ].
- Press [ENTER].
Example \(\PageIndex{2}\): Finding value of a common Logarithm using a Calculator
Evaluate \(y=\log (321)\) to four decimal places using a calculator.
Solution
- Press [LOG] .
- Enter 321 , followed by [ ) ] .
- Press [ENTER] .
Rounding to four decimal places, \(\log (321) \approx 2.5065\).
Analysis
Note that \(10^{2}=100\) and that \(10^{3}=1000\). Since 321 is between 100 and 1000, we know that \(\log (321)\) must be between \(\log (100)\) and \(\log (1000)\). This gives us the following:
\[\begin{array}{l}
100<321<1000 \\
2<205065<3
\end{array} \nonumber \]
Try it Now 5
Evaluate \(y = \log(123)\) to four decimal places using a calculator.
Example \(\PageIndex{3}\)
The 1994 Northridge Earthquake in Southern California registered a 6.7 on the Richter scale. Exactly a year later, the 1995 Kobe Earthquake in Japan registered a 6.9. Compare the amount of energy released from these two quakes.
Solution
\[10^{6.9-6.7}=10^{0.2}=1.585 \nonumber \]
The Kobe Earthquake released about 1.56 times the energy (about 56% more) released by the Northridge quake.
Example \(\PageIndex{4}\): Rewriting and solving a real-world exponential model
The amount of energy released from one earthquake was 500 times greater than the amount of energy released from another. The equation \(10^x=500\) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
Solution
We begin by rewriting the exponential equation in logarithmic form.
\[\begin{aligned}
10^{x} &= 500\\
\log (500) &= x \quad \text {Use the definition of the common log.}
\end{aligned} \nonumber \]
Next we evaluate the logarithm using a calculator:
- Press [LOG] .
- Enter 500, followed by [ ) ] .
- Press [ENTER] .
- To the nearest thousandth, \(\log (500) \approx 2.699\).
The difference in magnitudes was about 2.699. This means that the bigger earthquake has the magnitude about 2.7 more than the magnitude of the smaller earthquake.
Try it Now 6
The amount of energy released from one earthquake was 8,500 times as great as the amount of energy released from another. The equation \(10^x=8500\) represents this situation, where \(x\) is the difference in magnitudes on the Richter Scale. To the nearest thousandth, what was the difference in magnitudes?
Reference
- References (14)
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0


7.2C: Size Variation and ORF Contents in Genomes
- Explain prokaryotic genome size variation and ORFs
In molecular genetics, an open reading frame (ORF) is the part of a reading frame that contains no stop codons. The transcription termination pause site is located after the ORF, beyond the translation stop codon, because if transcription were to cease before the stop codon, an incomplete protein would be made during translation.
Normally, inserts which interrupt the reading frame of a subsequent region after the start codon cause frameshift mutation of the sequence and dislocate the sequences for stop codons.
Open reading frames are used as one piece of evidence to assist in gene prediction. Long ORFs are often used, along with other evidence, to initially identify candidate protein coding regions in a DNA sequence. The presence of an ORF does not necessarily mean that the region is ever translated. For example, in a randomly generated DNA sequence with an equal percentage of each nucleotide, a stop-codon would be expected once every 21 codons. A simple gene prediction algorithm for prokaryotes might look for a start codon followed by an open reading frame that is long enough to encode a typical protein, where the codon usage of that region matches the frequency characteristic for the given organism ‘s coding regions. Even a long open reading frame by itself is not conclusive evidence for the presence of a gene.
If a portion of a genome has been sequenced (e.g. 5′-ATCTAAAATGGGTGCC-3′), ORFs can be located by examining each of the three possible reading frames on each strand. In this sequence two out of three possible reading frames are entirely open, meaning that they do not contain a stop codon:
…A TCT AAA ATG GGT GCC…
…AT CTA AAA TGG GTG CC…
…ATC TAA AAT GGG TGC C…
Possible stop codons in DNA are “TGA”, “TAA”, and “TAG”. Thus, the last reading frame in this example contains a stop codon (TAA), unlike the first two.
Bacterial genomes display variation in size, even among strains of the same species. These microorganisms have very little noncoding or repetitive DNA, as the variation in their genome size usually reflects differences in gene repertoire. Some species, particularly bacterial parasites and symbionts, have undergone massive genome reduction and simply contain a subset of the genes present in their ancestors.
However, in free-living bacteria, such gene loss cannot explain the observed disparities in genome size because ancestral genomes would have had to contain improbably large numbers of genes. Surprisingly, a substantial fraction of the difference in gene contents in free-living bacteria is due to the presence of ORFans, that is, open reading frames (ORFs) that have no known homologs and are consequently of no known function.
The high numbers of ORFans in bacterial genomes indicate that, with the exception of those species with highly reduced genomes, much of the observed diversity in gene inventories does not result from either the loss of ancestral genes or the transfer from well-characterized organisms (processes that result in a patchy distribution of orthologs but not in unique genes) or from recent duplications (which would likely yield homologs within the same or closely related genome).
Key Points
- Open reading frames are used as one piece of evidence to assist in gene prediction.
- If a portion of a genome has been sequenced, ORFs can be located by examining each of the three possible reading frames on each strand.
- Bacterial genomes display variation in size, even among strains of the same species.
Key Terms
- gene : A unit of heredity; a segment of DNA or RNA that is transmitted from one generation to the next. It carries genetic information such as the sequence of amino acids for a protein.
- codons : The genetic code is the set of rules by which information encoded within genetic material (DNA or mRNA sequences) is translated into proteins (amino acid sequences) by living cells. Biological decoding is accomplished by the ribosome, which links amino acids in an order specified by mRNA, using transfer RNA (tRNA) molecules to carry amino acids and to read the mRNA three nucleotides at a time. The genetic code is highly similar among all organisms, and can be expressed in a simple table with 64 entries.
- open reading frame : A sequence of DNA triplets, between the initiator and terminator codons, that can be transcribed into mRNA and later translated into protein.


6.5: Administering Medications Rectally and Vaginally
Medication Administered Rectally
Drugs administered PR have a faster action than via the oral route and a higher bio-availability – that is, the amount of effective drug that is available is greater as it has not been influenced by upper gastrointestinal tract digestive processes. Rectal absorption results in more of the drug reaching the systemic circulation with less alteration on route. As well as being a more effective route for delivering medication, rectal administration also reduces side-effects of some drugs, such as gastric irritation, nausea and vomiting (Lowry, 2016, para 2). Rectal medications are given for their local effects in the gastrointestinal system (e.g., laxatives) or their systemic effects (e.g., analgesics when oral route is contraindicated). Rectal medications are contraindicated after rectal or bowel surgery, with rectal bleeding or prolapse, and with low platelet counts. Checklist 46 outlines the procedure for administering rectal suppositories or enemas.
Checklist 46: Medication Administered RectallyDisclaimer: Always review and follow your hospital policy regarding this specific skill. |
|||
Safety considerations:
|
|||
Steps |
Additional Information |
||
| 1. Check MAR against doctor’s orders. |
Check that MAR and doctor’s orders are consistent. Night staff usually complete and verify this check as well. |
||
2. Perform the SEVEN RIGHTS x 3 (must be done with each individual medication):
Medication calculation: D/H x S = A ( D or d esired dosage/ H or h ave available x S or s tock = A or a mount prepared) |
The right patient: check that you have the correct patient using two patient identifiers (e.g., name and date of birth).
The right medication (drug): check that you have the correct medication and that it is appropriate for the patient in the current context. The right dose: check that the dose makes sense for the age, size, and condition of the patient. Different dosages may be indicated for different conditions. The right route: check that the route is appropriate for the patient’s current condition. The right time: adhere to the prescribed dose and schedule. The right reason: check that the patient is receiving the medication for the appropriate reason. The right documentation: always verify any unclear or inaccurate documentation prior to administering medications. NEVER document that you have given a medication until you have actually administered it. |
||
3. The label on the medication must be checked for name, dose, and route, and compared with the MAR at three different times:
|
These checks are done before administering the medication to your patient. If taking drug to bedside (e.g., eye drops), do third check at bedside. |
||
| 4. If possible, have patient defecate prior to rectal medication administration. | Medication should not be inserted into feces. | ||
| 5. Ensure that you have water-soluble lubricant available for medication administration. | Lubricant reduces friction as suppository enters rectal canal. | ||
| 6. Explain the procedure to the patient. If patient prefers to self-administer the suppository/enema, give specific instructions to patient on correct procedure. | Patient may feel more comfortable self-administering suppository. | ||
| NOTE : Unintended vagal stimulation may occur, resulting in bradycardia in some patients. Be aware that the rectal route may not be suitable for certain cardiac conditions. Notify physician. | |||
7. Raise bed to working height.
|
Positioning helps prevent injury to nurse administering medication. This protects patient’s privacy and facilitates relaxation.
Drape protects linens from potential fecal drainage. |
||
| 8. Apply clean non-sterile gloves. | Gloves protect the nurse from contact with mucous membranes and body fluids. | ||
| 9. Assess patient for diarrhea or active rectal bleeding. | Rectal medications are contraindicated in these situations. | ||
| 10. Apply clean non-sterile gloves if previous gloves were soiled. | Gloves protect the nurse from contact with mucous membranes and body fluids. | ||
|
11. Remove wrapper from suppository/tip of enema and lubricate rounded tip of suppository and index finger of dominant hand with lubricant.
If enema, lubricate only tip of enema. |
Lubricant reduces friction as suppository/enema enters rectal canal. | ||
|
12. Separate buttocks with non-dominant hand and, using gloved index finger of dominant hand, insert suppository (rounded tip toward patient) into rectum toward umbilicus while having patient take a deep breath, exhale through the mouth, and relax anal sphincter.
If enema: Expel air from enema and then insert tip of enema into rectum toward umbilicus while having patient take a deep breath, exhale through the mouth, and relax anal sphincter. |
You should feel the anal sphincter close around your finger after insertion. Forcing the suppository/enema through a clenched sphincter will cause pain and, potentially, rectal damage. | ||
|
13. With your gloved finger, insert suppository along wall of rectum about 5 cm beyond anal sphincter. Do not insert the suppository into feces.
If enema: roll plastic bottle from bottom to tip until all solution has entered rectum and colon. |
Suppository should be against rectal mucosa for absorption and therapeutic action. Inserting suppository into feces will decrease its effectiveness. | ||
| 14. Option: A suppository may be given through a colostomy (not ileostomy) if prescribed. | The patient should lie supine and a small amount of lubricant should be used. | ||
| 15. Remove finger and wipe patient’s anal area. | Wiping removes excess lubricant and provides comfort to the patient. | ||
| 16. Ask patient to remain on side for 5 to 10 minutes. | This position helps prevent the expulsion of suppository. | ||
| 17. Discard gloves by turning them inside out and disposing of them and any used supplies as per agency policy. Perform hand hygiene. | Using gloves reduces transfer of microorganisms. | ||
| 18. Ensure call bell is nearby and bedpan/commode is available and close by. | If suppository is a laxative or stool softener, patient will require a bedpan/commode or close proximity to toilet. | ||
| 19. Document procedure as per agency policy and include patient’s tolerance of administration. | Timely and accurate documentation promotes patient safety. | ||
| Data source: BCIT, 2015; Lilley et al., 2011; Perry et al., 2014 |
Medication Administered Vaginally
Female patients may require vaginal suppositories to treat vaginal infections. Vaginal suppositories are larger and more oval than rectal suppositories, and are inserted with an applicator (see Figure 6.2) or by hand (see Figure 6.3). Checklist 47 outlines the procedure for administering vaginal suppositories or medications.
Checklist 47: Medication Administered VaginallyDisclaimer: Always review and follow your hospital policy regarding this specific skill. |
||||
Safety considerations:
|
||||
Steps |
Additional Information |
|||
| 1. Check MAR against doctor’s orders. |
Students must check that MAR and doctor’s orders are consistent.
Night staff usually complete and verify this check as well. |
|||
2. Perform the SEVEN RIGHTS x 3 (must be done with each individual medication):
Medication calculation: D/H x S = A ( D or d esired dosage/ H or h ave available x S or s tock = A or a mount prepared) |
The right patient: check that you have the correct patient using two patient identifiers (e.g., name and date of birth).
The right medication (drug): check that you have the correct medication and that it is appropriate for the patient in the current context. The right dose: check that the dose makes sense for the age, size, and condition of the patient. Different dosages may be indicated for different conditions. The right route: check that the route is appropriate for the patient’s current condition. The right time: adhere to the prescribed dose and schedule. The right reason: check that the patient is receiving the medication for the appropriate reason. The right documentation: always verify any unclear or inaccurate documentation prior to administering medications. NEVER document that you have given a medication until you have actually administered it. |
|||
3. The label on the medication must be checked for name, dose, and route, and compared with the MAR at three different times:
|
These checks are done before administering the medication to your patient. If taking drug to bedside (e.g., eye drops), do third check at bedside. |
|||
| 4. Before inserting the medication vaginally, explain the procedure to the patient. If patient prefers to self-administer the vaginal medication, give specific instructions to patient on correct procedure. | Patient may feel more comfortable self-administering vaginal medication. | |||
| 5. Ensure that you have water-soluble lubricant available for medication administration. | Lubricant reduces friction against vaginal mucosa as medication is inserted. | |||
| 6. Have patient void prior to procedure. | Voiding prevents passing of urine during procedure. | |||
7. Raise bed to working height.
|
Position helps prevent injury to nurse administering medication.
Draping protects patient’s privacy and facilitates relaxation. |
|||
| 8. Apply clean non-sterile gloves. | Gloves protect the nurse from contact with mucous membranes and body fluids. | |||
| 9. Remove suppository from wrapper and apply a liberal amount of water-soluble lubricant to suppository and index finger of dominant hand. Suppository should be at room temperature. | Lubricant reduces friction against vaginal mucosa as medication is inserted. | |||
| 10. With non-dominant hand, gently separate labial folds. With gloved index finger of dominant hand, insert lubricated suppository about 8 to 10 cm along posterior vagina wall. | Exposes vaginal orifice and helps to ensure equal distribution of medication. | |||
| 11. Withdraw finger and wipe away excess lubricant. | Wiping maintains patient comfort. | |||
| NOTE: An applicator may be used to insert vaginal medication. Follow procedure above and specific manufacturer directions. | ||||
| 12. Discard gloves by turning them inside out and disposing of them and any used supplies as per agency policy. Perform hand hygiene. | Using gloves reduces transfer of microorganisms. | |||
| 13. Document procedure as per agency policy, and include patient’s tolerance of administration. | Timely and accurate documentation promotes patient safety. | |||
| Data source: Lilley et al., 2011; Perry et al., 2014 |
Critical Thinking Exercises
- Your patient has a colostomy, and a laxative has been prescribed. Discuss the procedure for administering a laxative in this situation.
- Your patient prefers to self-administer her vaginal suppository. Outline the steps you would explain for safe and appropriate administration of a vaginal medication.


15.3: Hydrodynamics
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In the previous sections we developed “hydrostatic” models for fluids when those fluids are at rest (in some inertial reference frame). In this section, we develop “hydrodynamic” models to discuss what happens when fluids flow. We will restrict our models to fluids that flow in a “laminar” fashion, rather than a “turbulent” fashion.
Laminar flow is the flow of a fluid when each particle in the fluid follows a path that can be represented by a line (a “streamline”). Turbulent flow is the flow of a fluid where particles can follow rather complex paths, usually involving “Eddy currents” (little whirlpools). The two types of flow are illustrated in Figure \(\PageIndex{1}\).
Continuity of flow
Consider the laminar flow of a fluid through a pipe whose cross-sectional area narrows from \(A_{1}\) to \(A_{2}\) in the direction of flow, as illustrated in Figure \(\PageIndex{2}\).
The particles that make up the fluid have a speed \(v_{1}\) at the wide end of the pipe and speed \(v_{2}\) at the narrow end. The equation of continuity is based on the premise that the fluid that enters the pipe must exit the pipe, as there is nowhere else for the fluid to go. That is, if during a period of time, \(∆t\), a mass, \(∆m\), of fluid enters the wide end of the pipe, then during that same period of time, the same mass of fluid must exit the narrow end of the pipe.
During a period of time, \(∆t\), the fluid at the wide end of the pipe will travel a distance \(l_{1} = v_{1}∆t\). Thus, a volume of fluid, \(∆V_{1}\), will enter the wide end of the pipe:
\[\begin{aligned} \Delta V_{1}=A_{1}l_{1}=A_{1}v_{1}\Delta t \end{aligned}\]
Similarly, during that period of time, a volume \(∆V_{2}\) will exit the narrow end of the pipe:
\[\begin{aligned} \Delta V_{2}=A_{2}l_{2}=A_{2}v_{2}\Delta t \end{aligned}\]
If the fluid is compressible, its density can change. Let \(ρ_{1}\) be the density of the fluid at the wide end of the pipe and \(ρ_{2}\) be the density of the fluid at the narrow end. The mass of fluid, \(∆m\), entering the wide end of the pipe is given by:
\[\begin{aligned} \Delta m = \rho _{1}\Delta V_{1}=\rho _{1}A_{1}v_{1}\Delta t \end{aligned}\]
The mass of fluid exiting the narrow end of the pipe is given by:
\[\begin{aligned} \Delta m = \rho _{2}\Delta V_{2} = \rho _{2}A_{2}v_{2}\Delta t\end{aligned}\]
The mass of fluid entering the wide end of the pipe must equal the mass exiting the narrow end of the pipe:
\[\begin{aligned} \rho _{1}A_{1}v_{1}\Delta t = \rho _{2}A_{2}v_{2}\Delta t \end{aligned}\]
Leading to the equation of continuity:
\[\rho _{1}A_{1}v_{1}=\rho _{2}A_{2}v_{2}\]
The quantity \(ρAv\) has dimensions of mass per time, and corresponds to the mass of fluid passing through a cross section \(A\) per unit time.
If the fluid is incompressible, as are most liquids, then the density is the same on both sides of the pipe, and the equation simplifies to:
\[A_{1}v_{1}=A_{2}v_{2}\qquad\text{(Incompressible fluid)}\]
For a liquid, we can define the “volumetric flow”, \(Q\), as:
\[\begin{aligned} Q=Av \end{aligned}\]
where \(A\) is the cross-sectional area of the surface through which a fluid with speed, \(v\), flows 1 . \(Q\) has the dimension of volume per time, and corresponds to the volume of fluid passing through the cross section \(A\) per unit time. For an incompressible fluid, the equation of continuity is thus equivalent to stating that the volumetric flow, \(Q\), of the fluid is a constant.
Exercise \(\PageIndex{1}\)
When water flows out of your faucet, you observe that the stream of water gets narrower as the water moves down, as shown in Figure \(\PageIndex{3}\). Why is this?
- The atmospheric pressure increases as the waters moves downwards, so the stream of water is more and more compressed.
- As the water accelerates due to gravity, the cross-sectional area of the flowing water must reduce in order to preserve a constant flow rate.
- Answer
- B.
Example \(\PageIndex{1}\)
Your garden hose has a diameter of \(D = 2\text{ cm}\). How fast must water flow out of the hose if you are to fill a \(5\text{ l}\) bucket in one minute?
Solution
We need the volume flow rate from the hose to be \(Q = 5\text{ l/min}\). We can convert this to SI units:
\[\begin{aligned} Q=(5\text{l/min})\left( \frac{1}{1000}\text{m}^{3}\text{/l} \right)\left( \frac{1}{60}\text{min/s} \right)=\frac{5}{6\times 10^{4}}\text{m}^{3}\text{/s} = 8.3\times 10^{-5}\text{m}^{3}\text{/s} \end{aligned}\]
Since we know the area of the hose, we can determine the speed of the water to achieve the given flow rate:
\[\begin{aligned} Q&=Av=\pi \left(\frac{D}{2} \right)^{2}v \\[4pt] \therefore v&=\frac{Q}{\pi \left( \frac{D}{2} \right)^{2}}=\frac{(8.3\times 10^{-5}\text{m}^{3}\text{/s})}{\pi (0.01\text{m})^{2}}=0.265\text{m/s} \end{aligned}\]
Bernoulli's Principle
In this section, we examine how the pressure and speed of a fluid change as a the fluid flows. We will restrict ourselves to discussing the laminar flow of an incompressible fluid with no friction. Bernoulli was the first to quantitatively describe the flow of incompressible fluids, and we will show in this section how to derive “Bernoulli’s Principle”.
Consider the laminar flow of an incompressible fluid through a pipe that changes height, from \(y_{1}\) to \(y_{2}\), as well as cross-sectional area, from \(A_{1}\) to \(A_{2}\), as shown in Figure \(\PageIndex{4}\). The figure shows an element of fluid, in blue, as it moves through the pipe. The top panel corresponds to the location of the fluid element at time \(t = 0\), whereas the bottom panel shows the location of the element of fluid at time \(t = ∆t\).
To model how the fluid moves through this pipe, we can use energy and the Work-Energy Theorem. We start by considering the amount of work done on the element of fluid as it moves from the position in the top panel to the position in the bottom panel.
The fluid that is to the left of the element of fluid exerts a pressure, \(P_{1}\), on the fluid element that leads to a net force, \(\vec F_{1}\), towards the right. Similarly, the fluid to the right of the element of fluid exerts a net force \(\vec F_{2}\) in the opposite direction, due to the pressure \(P_{2}\) on that side of the fluid element.
In a period of time, \(∆t\), the left part of the fluid element will move a distance \(l_{1} = v_{1}∆t\), while the right part of the fluid element will move a distance \(l_{2} = v_{2}∆t\). We can calculate the work done by each force, defining positive work to be in the direction of motion:
\[\begin{aligned} W_{1}&=F_{1}l_{1}=(P_{1}A_{1})(v_{1}\Delta t) \\[4pt] W_{2} &=-F_{2}l_{2}=-(P_{2}A_{2})(v_{2}\Delta t) \end{aligned}\]
Gravity will also do (negative) work on the fluid as it changes height. In a period of time, \(∆t\), a mass of fluid, \(∆m\), will move from position \(y = y_{1}\) to position \(y = y_{2}\). The mass of fluid that changes height is given by the part of the fluid that moves a distance, \(l_{1}\), on the right side of the pipe:
\[\begin{aligned} \Delta m = V_{1}\rho = A_{1}l_{1}\rho = A_{1}v_{1}\Delta t\rho \end{aligned}\]
Because of the equation of continuity, this is also equal to the mass of fluid that moves a distance, \(l_{2}\), on the left side of the pipe:
\[\begin{aligned} \Delta m = V_{2}\rho = A_{2}l_{2}\rho = A_{2}v_{2}\Delta t\rho \end{aligned}\]
since \(A_{1}v_{1}=A_{2}v_{2}\).
The force of gravity will thus do negative work on that mass element:
\[\begin{aligned} W_{g}=-\Delta mg(y_{2}-y_{1})=-(A_{1}v_{1}\Delta t\rho)g(y_{2}-y_{1}) \end{aligned}\]
The net work done on the element of fluid over the time \(∆t\) is thus:
\[\begin{aligned} W^{net}=W_{1}+W_{2}+W_{g}=P_{1}A_{1}v_{1}\Delta t-P_{2}A_{2}v_{2}\Delta t - A_{1}v_{1}\Delta t \rho g(y_{2}-y_{1}) \end{aligned}\]
Note that, because of the equation of continuity, \(A_{1}v_{1} = A_{2}v_{2}\), we can factor out a \(A_{1}v_{1}\) from each term:
\[\begin{aligned} W^{net}=A_{1}v_{1}\Delta t(P_{1}-P_{2}-\rho g(y_{2}-y_{1})) \end{aligned}\]
The net work done on the fluid must equal the change in kinetic energy, \(∆K\), of the mass element, \(∆m\), from one end of the pipe to the other:
\[\begin{aligned} \begin{aligned} \Delta K&=\frac{1}{2}\Delta mv_{2}^{2} - \frac{1}{2}\Delta mv_{1}^{2} \\[4pt] &=\frac{1}{2}(A_{1}v_{1}\Delta t\rho )(v_{2}^{2}-v_{1}^{2}) \end{aligned} \end{aligned}\]
Using the Work-Energy Theorem, we have:
\[\begin{aligned} W^{net}&=\Delta K \\[4pt] A_{1}v_{1}\Delta t\left(P_{1} - P_{2}-\rho g(y_{2}-y_{1})\right)&=\frac{1}{2}(A_{1}v_{1}\Delta t\rho )(v_{2}^{2}-v_{1}^{2}) \\[4pt] P_{1}-P_{2}-\rho g(y_{2}-y_{1})&=\frac{1}{2}\rho v_{2}^{2}-\frac{1}{2}\rho v_{1}^{2} \end{aligned}\]
We can re-arrange this so that all the quantities for each side of the pipe are on the same side of the equation:
\[\begin{aligned} P_{1}+\frac{1}{2}\rho v_{1}^{2} +\rho gy_{1}=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2} \end{aligned}\]
Since the locations 1 and 2 that we chose are arbitrary, we can state that, for laminar incompressible flow, the following quantity evaluated at any position is a constant:
\[P+\frac{1}{2}\rho v^{2}+\rho gy = \text{constant}\]
This statement is what we call Bernoulli’s Equation, and is equivalent to conservation of energy for the fluid. If the fluid is not flowing \((v_{1} = v_{2} = 0)\), then this is equivalent to the statement of hydrostatic equilibrium that we derived in Equation 15.1.1 :
\[\begin{aligned} P_{1}+\rho gy_{1}&=P_{2}+\rho gy_{2} \\[4pt] \therefore P_{2}-P_{1}=-\rho g(y_{2}-y_{1}) \end{aligned}\]
If the flow of the fluid is at constant height \((y_{2} = y_{1})\), then Bernoulli’s equation can be written as:
\[\begin{aligned} P_{1}+\frac{1}{2}\rho v_{1}^{2}=P_{2}+\frac{1}{2}\rho v_{2}^{2} \end{aligned}\]
If a fluid is flowing at constant height such that \(v_{2} > v_{1}\) (as in Figure \(\PageIndex{2}\)), then \(P_{2} < P_{1}\); that is, the pressure in the fluid is lower if the fluid is flowing faster . Note that \(P\) is the pressure inside the fluid and is not related to the force that would be exerted by the fluid if it were to collide with an object. It makes sense that the fluid has a lower pressure where it is moving faster, because the net force exerted on the fluid is related to the difference in pressure on either side of the fluid. The fluid will accelerate in the direction where pressure decreases, thus it will be moving faster when it is in a region of low pressure.
Bernoulli’s principle can be used to describe many phenomena. For example, an airplane wing (technically, an “airfoil”) creates lift because the pressure of the air above the wing is lower than the pressure above the wing. This is illustrated in Figure \(\PageIndex{5}\), which shows that the laminar flow of the air creates a low pressure area above the wing. As the stream lines of air encounter the wing, those that are above the wing get compressed together, which leads to a faster speed of the air above the wing (equation of continuity). The resulting difference in air pressure above and below the wing results in a net upwards force on the wing.
Bernoulli’s principle also describes why the roof can be lifted off of a house in high winds (Figure \(\PageIndex{6}\), left panel). It is not the force of the wind against the roof that blows the roof off of a house; it is the difference in air pressure in the house (normal) and the pressure above the roof (low, due to the flowing wind), that results in a net upwards force on the roof. Bernoulli’s principle is also used to construct atomizers which allow liquid in a bottle to be sprayed (Figure \(\PageIndex{6}\), right panel). For example, perfume bottles often have a bulb connected to a tube/spout. When you squeeze the bulb, it causes the air in the tube to flow quickly, creating a low pressure in the vertical segment of the spout. The liquid is forced up by the pressure in the bottle; once the liquid arrives in the fast flowing air, it is sprayed out along with the air.
Exercise \(\PageIndex{2}\)
When a high speed train is traveling at constant speed, is there a net force on the windows from air pressure?
- No, since the windows are stationary relative to the train, there is no net force on them from air pressure.
- Yes, there is a net outwards force on the windows from air pressure.
- Yes, there is a net inwards force on the windows from air pressure.
- Answer
- B.
The following examples illustrate how to apply Bernoulli’s principle.
Example \(\PageIndex{2}\)
A water tower is constructed so that the bottom of the water tank is a height \(h_{2}\) above the ground, as illustrated in Figure \(\PageIndex{7}\). The water in the tank is at a height \(h_{1}\) from the bottom of the tank. A leak from a hole is found at the base of the tank (the water flows horizontally out of the hole). What is the horizontal distance, \(d\), from the bottom of the tower to where the water from the leak hits the ground? Assume that the water level in the tank is constant and that atmospheric pressure does not change appreciably over the height of the tower.
Solution
The pressure in the water tank leads to the water exiting the bottom of the tank with a horizontal velocity of magnitude, \(v\). That water then undergoes projectile motion on its way to the ground. We can first determine the speed of the water exiting the tank and then use the kinematics for projectile motion to model the distance, \(d\).
We model the flow of the water using a two-dimensional coordinate system with a horizontal \(x\) axis (positive to the right), and a vertical \(y\) axis (positive upwards). We place the origin at the bottom of the water tower, on the ground, below the hole, as shown in Figure \(\PageIndex{7}\).
At the top of tank, at a height \(y = h_{1} + h_{2}\), the water has a speed of zero and is at atmospheric pressure, \(P_{0}\). At the exit of the hole at the bottom of the tank, at a height \(y = h_{2}\), the water has a speed \(v_{2}\) and is also at atmospheric pressure. Using Bernoulli’s equation at the top (1) and bottom (2) of the tank, we have:
\[\begin{aligned} P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}&=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2} \\[4pt] P_{0} +(0) +\rho g(h_{1}+h_{2})&=P_{0}+\frac{1}{2}\rho v^{2}+\rho gh_{2} \\[4pt] \therefore v_{2}&=\sqrt{2gh_{1}} \end{aligned}\]
which is exactly the speed that any object falling a distance \(h_{1}\) would have.
Using kinematics, we can find the time that it would take the water to fall a distance \(h_{2}\) (where the water’s velocity is horizontal as it exits the tank):
\[\begin{aligned} h_{2}&=\frac{1}{2}gt^{2} \\[4pt] \therefore t&=\sqrt{\frac{2h_{2}}{g}} \end{aligned}\]
The distance \(d\) covered by the water is thus given by:
\[\begin{aligned} d=v_{2}t=\sqrt{2gh_{1}}\sqrt{\frac{2h_{2}}{g}}=\sqrt{4h_{1}h_{2}} \end{aligned}\]
Discussion
We find that the water coming out of the bottom of the tank, when there is a height, \(h_{1}\), of water above it providing pressure, will have the same speed as that of a particle which has fallen a distance, \(h_{1}\). This is because there is no net pressure difference between the top of the water tank and where the water has exited the hole, so gravity is the only force doing work on the water. Gravity will do work at the same rate on particles of water as on any other particle, so the speed of the water particles at the bottom of the tank is the same as if they had fallen a distance, \(h_{1}\). Again, once the water particles are falling through the air, gravity is the only net force exerted on those particles, so they undergo projectile motion, just as any other particle would.
Example \(\PageIndex{3}\)
You measure that water comes out of your kitchen faucet at a rate of \(6\text{ l/min}\). The faucet has a diameter of \(2\text{ cm}\). At what rate will water flow out of your basement faucet, which has a diameter of \(1\text{ cm}\) and is located a height, \(h=\text{ 3 m}\), below your kitchen faucet? Assume that atmospheric pressure, \(P_{0}\), does not change appreciably between your kitchen and basement.
Solution
The water flows out of the kitchen faucet at a speed, \(v_{1}\), where the pressure is atmospheric. If the area of the kitchen faucet is \(A_{1}\) we can determine the speed, \(v_{1}\), from the given flow rate, \(Q_{1} = 6 \text{l/min} = 1 × 10^{−4} \text{m}^{3}\text{/s}\):
\[\begin{aligned} Q_{1}&=A_{1}v_{1} \\[4pt] v_{1}&=\frac{Q_{1}}{A_{1}}=\frac{(1\times 10^{-4}\text{m}^{3}\text{/s})}{\pi (0.01\text{cm})^{2}}=0.318\text{m/s} \end{aligned}\]
The water will flow out of the basement faucet with a speed, \(v_{2}\), where the pressure is also atmospheric, \(P_{0}\). We can use Bernoulli’s equation to relate the flow out of the basement faucet (2) to that at the kitchen faucet (1). We choose the \(y\) axis of a vertical coordinate system such that the basement is located at \(y_{2} = 0\) and the kitchen faucet is located at \(y_{1} = 3\text{ m}\):
\[\begin{aligned} P_{1}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}&=P_{2}+\frac{1}{2}\rho v_{2}^{2}+\rho gy_{2} \\[4pt] P_{0}+\frac{1}{2}\rho v_{1}^{2}+\rho gy_{1}&=P_{0}+\frac{1}{2}\rho v_{2}^{2} \\[4pt] \frac{1}{2}v_{1}^{2}+gy_{1}&=\frac{1}{2}v_{2}^{2} \\[4pt] \therefore v_{2}&=\sqrt{v_{1}^{2}+2gy_{1}} \\[4pt] &=\sqrt{(0.318\text{m/s})^{2}+2(9.8\text{m/s}^{2})(3\text{m})} = 7.67\text{m/s} \end{aligned}\]
The corresponding flow rate at the basement faucet will be:
\[\begin{aligned}Q_{2}=A_{2}v_{2}=\pi(0.005\text{m})^{2}(7.67\text{m/s})=6.03\times 10^{-4}\text{m}^{3}\text{/s}=36.171\text{/min}\end{aligned}\]
Discussion
We find that the flow rate out of the basement faucet is six times that at the kitchen faucet. The speed of the water coming out of the basement faucet is more than \(20\) times the speed of the water at the kitchen faucet. Although it is true that one gets better water pressure out of a faucet that is lower in the building, this change in flow is unrealistically high, and this is a poor model for flow of water in the pipes of your house.
You can easily verify that the speed of the water in different levels of your house does not vary by a factor near \(20\) for a \(3\text{ m}\) change in height (you could compare the flow rate for two faucets with the same diameter). This is because our model neglects the effect of friction as water flows in the pipes; in reality, there is much greater pressure in the pipes than that due to gravity, as well as a gradient in the pressure in your pipes, that will lead to the flow rates being similar in your kitchen and basement.
Viscosity
So far, we have assumed that fluids flow with no friction. In reality, the particles moving in a fluid exert internal friction on each other called “viscosity”. This can be modeled as the friction between different layers of fluid in a laminar flow. For example, you may notice that the water that flows in a wide river flows much faster in the middle of the river than near the river banks, where the water is almost stationary, as shown in Figure \(\PageIndex{8}\).
One can model the banks of the river as exerting a frictional force on the layer of water that is in contact with the banks. That layer then exerts a frictional force on the next layer closer to the center of the river, and so on.
One can define a viscosity coefficient, \(η\), based on measuring the force required to pull a plate past another plate when there is a fluid between the plates. Consider two plates that have an area, \(A\), that are a distance \(l\) apart, and contain the fluid of interest between them, as illustrated in Figure \(\PageIndex{9}\).
The viscosity of the fluid is defined based on the force that is required to pull the top plate while the bottom plate remains immobile. The layer of fluid directly below the moving plate will move with the plate at a speed, \(v\), while the layer of fluid immediately in contact will the stationary plate will also be stationary. Moving one plate will thus lead to a gradient (a change) in the speed of the fluid as a function of the position between the two plates. The magnitude of the force, \(\vec F\), required to move one plate with speed, \(v\), was empirically determined to be proportional to the area of the plates, \(A\), and the speed, \(v\), while being inversely proportional to the distance, \(l\), between the two plates:
\[\begin{aligned} F ∝ A\frac{v}{l} \end{aligned}\]
The constant of proportionality is defined as the viscosity, \(η\), of the fluid:
\[F=ηA\frac{v}{l}\]
If the viscosity of the fluid is zero, then no force is required to pull the plate. The more viscous the fluid, the more difficult it is to pull the top plate. You can experiment with this by comparing the force required to move a small piece of paper across the top of a puddle of water and across the top of honey.
The presence of viscosity means that any fluid that flows will lose mechanical energy due to internal friction (which will heat up the fluid). Thus, Bernoulli’s equation is not correct if the fluid has viscosity, as a fluid cannot flow through a horizontal pipe without a change in pressure to overcome the losses due to friction.
Poiseuille Flow
For the flow of an incompressible viscous fluid through a pipe, one can postulate that the flow rate, \(Q\), is proportional to the change in pressure, \(∆P\), across the pipe:
\[\begin{aligned} Q ∝ ∆P \end{aligned}\]
where \(∆P\) is taken as the positive difference between the pressure at either end of the pipe. The fluid flows from high pressure to low pressure. We can introduce a constant of proportionality, \(R\), to be the “resistance of the pipe”, so that we can write:
\[\begin{aligned} Q=\frac{\Delta P}{R} \end{aligned}\]
where we wrote the constant of proportionality as \(1/R\), so that a larger value of \(R\) corresponds to a pipe with a higher resistance to flow. That is, for a given pressure difference, as one increases the resistance of the pipe, one decreases the flow rate through that pipe. The relationship above can be used to empirically determine the resistance of a pipe.
The flow through a pipe with a given resistance will be zero if there is no pressure gradient in the fluid along the pipe. Conversely, if there is no flow of fluid in the pipe, the pressure is the same everywhere in the pipe. We can thus also view a drop in pressure in a pipe to be the result of flow of liquid through the pipe. The pressure cannot drop across a horizontal pipe if there is no flow.
When you close the tap on your kitchen faucet, the pressure inside the faucet is close to the pressure in the main water line that supplies your house. As soon as you open the tap and allow water to flow, the pressure in your faucet drops to atmospheric pressure, and the resulting pressure gradient from the main supply forces water to flow out of the faucet. If you try to plug your kitchen faucet with your thumb and stop the flow of water, you will need to exert a force large enough to overcome the pressure that exists in the main water supply. You will find that it is practically impossible to stop the flow of water with your thumb, as the pressure in the main supply needs to be high enough to overcome the resistance of the pipes and still result in a usable flow of water.
Poiseuille first developed a model for the laminar flow of a liquid through a uniform horizontal cylindrical pipe of length, \(L\), with a circular cross-section with radius \(r\). He found that the resistance of such a pipe to a fluid of viscosity, \(η\), is given by:
\[\begin{aligned} R=\frac{8ηL}{\pi r^{4}} \end{aligned}\]
This makes some intuitive sense, as we expect more resistance (more impedance to flow), if the pipe is longer and if the fluid is more viscous (the resistance is zero if there is no viscosity). We further expect less resistance if the pipe has a larger radius. The resistance found by Poiseuille goes down as the fourth power of the radius. Thus, a pipe that is twice as wide will have a volume flow that is \(2^{4} = 16\) times larger because of the reduced resistance.
The laminar flow rate, \(Q\), of a viscous fluid through a pipe of length \(L\) and radius \(R\), when there is a pressure difference \(∆P\), is given by:
\[Q=\frac{\pi r^{4}}{8ηL}\Delta P\]
This is usually referred to as “Poiseuille’s Equation”.
Exercise \(\PageIndex{3}\)
Does the flow rate of water out of a garden hose depend on the length of the hose?
- No, since the volume of water entering the hose must also exit the hose, it does not matter how long the hose is.
- Yes, the resistance of the hose depends on its length, so the pressure drop across the hose will change, and so will the flow rate.
- Answer
- B.
Example \(\PageIndex{4}\)
You are modeling the flow of water for a city. Two houses are connected in parallel to the main water supply, so that water from the main supply flows into either house 1 or house 2, and the flows out of each house then join up again at the main supply. The difference in pressure, \(∆P\), between the entry and exit point of water is the same for each house, and each house can be modeled as having a net resistance, \(R_{1}\) or \(R_{2}\), to the flow of water, as illustrated in Figure \(\PageIndex{10}\). If you model the two houses as being the equivalent of a single “effective” house with an effective resistance \(R\), what is the value of \(R\) in terms of \(R_{1}\) and \(R_{2}\)?
Solution
The water from the main will have to flow through either house 1 or house 2. If the flow rate through the main is \(Q\), we require that this be equal to the sum of the flow rates through each house:
\[\begin{aligned} Q=Q_{1}+Q_{2} \end{aligned}\]
The flow through each house is related to the pressure difference, \(∆P\), across each house (which is the same), as well as the resistance of that house:
\[\begin{aligned} Q_{1}&=\frac{\Delta P}{R_{1}} \\[4pt] Q_{2}&=\frac{\Delta P}{R_{2}} \end{aligned}\]
The total flow rate is thus:
\[\begin{aligned} Q=Q_{1}+Q_{2}&=\frac{\Delta P}{R_{1}}+\frac{\Delta P}{R_{2}} \\[4pt] &=\Delta P \left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) \end{aligned}\]
We can write this as the flow through an effective resistance, \(R\), with a pressure difference \(∆P\):
\[\begin{aligned} Q&=\frac{\Delta P}{R} \\[4pt] \therefore R&=\frac{1}{\frac{1}{R_{1}}+\frac{1}{R_{2}}} \end{aligned}\]
Discussion
By requiring that the sum of the flows of water through the houses be the same as the flow rate through the main pipe, we were able to model the two houses as a single effective house with resistance \(R\). You may notice that this is the same relation as the equivalent resistance for two electrical resistors combined in parallel. This is because the flow of electrical current in a resistor can be modeled using similar tools to those required for modeling the flow of a viscous fluid in a pipe.
Footnotes
1. If the velocity of the fluid is not perpendicular to the surface, then \(v\) is the component of the velocity perpendicular to the surface.


3.1: Prelude to Trigonometric Identities and Equations
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Math is everywhere, even in places we might not immediately recognize. For example, mathematical relationships describe the transmission of images, light, and sound. The sinusoidal graph in Figure \(\PageIndex{1}\) models music playing on a phone, radio, or computer. Such graphs are described using trigonometric equations and functions. In this chapter, we discuss how to manipulate trigonometric equations algebraically by applying various formulas and trigonometric identities. We will also investigate some of the ways that trigonometric equations are used to model real-life phenomena.
Figure \(\PageIndex{1}\): A sine wave models disturbance. (credit: modification of work by Mikael Altemark, Flickr).


3.1: Introduction
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- John T. Caldwell
- University of California Press
In the heady air of an MIT Transmedia conference, the “aca-pro” audience voiced appreciation as the futurist digital media consultant bragged about how nonhierarchical innovation hot spots like the one he’d created in his boutique company were poised to make old, conservative approaches to film and television production obsolete. Like dinosaurs and “Detroit,” he argued, lazy, inefficient “old media” film/TV production professionals—who, like the auto industry, had lived long past their prime—could vanish and no tears would be shed. The unequivocal message: good riddance. Another panelist, an edgy new media branding consultant, sketched out some of his own recent viral marketing and stealth stunts that had successfully created “buzz” while costing the client little money. One stealth stunt involved triggering the Los Angeles Police Department, law enforcement helicopters, and public first responders to hover around a fake emergency. News coverage of this fake “media event” indeed spilled onto the marketer’s covert goal: greater notoriety for a transmedia start-up in Hollywood. Again, the MIT audience knowingly giggled at the sophisticated ironies in tricking tax-supported public infrastructure to unknowingly provide the “free” heavy marketing muscle required to launch a bit of edgy new intellectual property (IP). No one, however, discussed the political-economic or ethical downsides that this stunt buzz-making involved. Who were these people, both the aca-pro panelists and conference attendees, I wondered? How were they paid, and by whom , and for what , exactly? Cultural geography might provide the answers. Most of the visionaries were from New York or Boston (not Detroit or Los Angeles), where creative workers apparently no longer need or want to be paid, or have benefits, like the dinosaur film/TV/auto workers out west, mired as they were in the outdated heavy-industry quagmires apparently entombing them.
And why was I at this conference, given that the celebrated viral marketing “innovations” and free labor being worshipfully gossiped about here would horrify the fieldwork informants that I had been talking to: professional cinematographers, editors, directors, and grips? Of course, like some of the panelists, I had been publishing on “convergence media,” “repurposing,” and programming though “content migration” for some time. But my understanding of these current new media practices now seemed—from the perspective of Cambridge—to have come from some distant planet rather than the clean, cost-free world being celebrated at MIT. Then it hit me. My conference trip to Cambridge involved time travel; I’d fallen back thirty years into art school, and these capitalist marketing executives had become the new avant-garde: conceptual artists, performance artists, street artists, and provocateurs. But unlike their 1960s and 70s predecessors from the art world, these new-media conceptual artists were now handsomely paid for their faux outsiderness, unruly marketing innovations, snark, and boundary-crossing provocations—while simultaneously being lauded for their bored and studied public disinterest in matters of wages, benefits, or job security. If transmedia and viral marketing and branding consultants were the new “conceptual artists” of the twenty-first century, then my research must be clinging to dying professional communities defined by something more archaic and suspect: “craft” (also known as the innovator’s “other”).
Based on this encounter, I’d like to begin with three simple and very basic questions, before taking on and unpacking the three terms in my chapter title. First, why does TV labor matter to media aesthetics or TV studies? Second, how can or should we study it, given widespread and disruptive recent changes? And finally, given those same disruptions, where does TV production actually exist anymore? That is, where and how do we meaningfully locate production for research in the digital era? These questions are particularly acute in the American media sectors within which I operate—where government regulation and funding have withered, where neoliberal economics dominates, where traditional producing arrangements have disintegrated, where online crowdsourcing (via Kickstarter or Indiegogo) has become a legitimate option even for the unapologetic high-level industry professionals who increasingly slum there.
The last of my three questions actually complicates the first two, so I’d like to start there. Two possible answers to the question of where production is located were offered by economic geographer Allen Scott, as well as political economist Toby Miller and his coauthors. 1 Targeting Hollywood, both rebuffed the common clichés about production—that “it is a state of mind”—but did so in different ways. Scott’s research on material resource agglomeration undercut the ephemerality state-of-mind cliché. His account detailed why certain film/television nexus points survive as geographical centers despite the clear economic advantages that might be gained by moving somewhere else. Miller and his coauthors, by contrast, disrupt the lie that geographic inertia or exceptionalism anchors production in any way, arguing that the real subject for production research today can be found in what they term the New International Division of Cultural Labor (NICL), which can migrate or shape-shift in response to rapid economic change.
Whereas Scott examines the regional anchoring of production and Miller the global dispersion and splintering of production through runaway production, my research leads me to suggest a third alternative. That is, our current predicament may follow from our failure to recognize that a widely dispersed conceptualizing process may be as central to the core of television/media production today as the industrial and material production of series, formats, and network programming once was (features that once garnered the lion’s share of attention from critics and media scholars). I am suggesting here that hybrid forms of imaginative/economic speculation now systematically animate media production. Speculation—or “spec work,” as I will call it—has become a fundamental part of the complex economies of TV. Figuring out how to manage spec work from the deregulated creative labor “herd” helps provide rationality for TV industries as they seek to master (and eventually monetize) the unstable world of unruly fans, digital media, and remix and gift economies.
In saying this, I am not reverting to Scott’s and Miller’s target—media as “a state of mind” cliché—since the dispersed conceptualizing process I am targeting is as much a result and defining property of contemporary media labor as are the onscreen series that TV labor officially produces. TV is more than just the end product of TV production labor. I take television labor to be anticipatory as well—to include the endless prototyping, brainstorming, work shopping, ad hoc viral repackaging, and vocational spinning that precede and follow the shows for which TV companies officially take credit. Significantly, anticipatory spec work adds economic value to TV shows even if TV producers and executives ignore it. I am especially sensitive to Mayer and Stahl’s critiques of labor “erasures.” 2 My books Televisuality and Production Culture were both premised on paying greater attention, even in aesthetic studies, to the cultural functions and institutional logics of physical production and creative workers—things that critical TV scholars had in many cases largely overlooked or dismissed. 3 Over the past two decades, I’ve also found that an entirely different work activity infuses physical production, one based on recurrent cognitive speculation about imagined, experiential, onscreen worlds of one sort or another. To clarify: I am not talking about the construction of “imagined narrative worlds” driven by fans in “transmedia franchises” of the sort Henry Jenkins has postulated. 4 I am, rather, talking about the commercial “labor” of habitual and calculated speculation now found in workaday, frequently unremarkable television job sites.
Significantly, spec work can be found in both below-the-line and above-the-line production sectors. Which means that conjecture about imagined worlds increasingly functions as part of lowly, run-of-the-mill trade practice. Anticipatory labor is not owned or triggered exclusively by the “creatives,” executives, and producers, and can be found as well in the lowly technical crafts. In rejecting the exceptionalism we normally assign or reserve for the creative higher-ups in TV—the showrunner, producer, director, or executive—I am only arguing that we need to augment what we research; that we take seriously the rich terrain of cultural conjecture and anticipatory expression that now makes up the below-the-line worker’s skill set. Such things now function as an integral part of the bigger system we think of as “television.” Spec work is both workmanlike and ubiquitous, rather than unique in any way, and thus challenges media studies to rethink the parameters and boundaries we assume in production labor research.
Before drilling down deeper to examine spec work as a practice, I must clarify that the shift toward habitualized speculation as craft/creative work on the micro level, which I have just described, is linked to bigger changes in the macroscopic market predicament and thus transnational goals of many production companies, studios, and networks today. Specifically, success in media markets today depends less and less on the fabrication of a durable distributable entertainment object—which historically was the basis for television’s core project of owning shows and syndicating series. In the old system, we thought that a production was over when we “locked” picture and soundtrack, then timed (or color-corrected) and archived a stable program master, and sold copies and versions in markets for distribution to buyers. Now the notion that our program masters are never done, always prone to change, goes well beyond the traditional alterations—remixing, recutting, dubbing—required for international distribution. Producers now know, upfront, that it is even possible (via corporate contracts as yet unknown) to completely recreate interior scenes through the digital imposition of new product placements, online links, and integrated sponsorship within preexisting narratives, onscreen. And this directly affects creative decisions creators make on the set. Digital renders masters completely malleable, reworkable, remakable, endlessly. These changes are not completely novel but a matter of degree, since we have always trimmed scenes for breaks to intercut ads, converted NTSC to PAL standards, panned-and-scanned or letter-boxed, and altered program masters for foreign languages when needed. What was once secondary is now primary, however, with masters now malleable not just at the level of plot or episode but at the level of the pixel, with effects layers “inside” fictional narratives and dramatic scenes as well. The growing presumption of an endlessly malleable program master means that the entire process of television production can now be imagined as an anticipatory function of postproduction, with the potential for (and goal of) an endless lucrative life on the “back end” of a project. Proliferating digital technologies mean that most forms of production can be understood as functions of postproduction—where the cognitive work of preproduction speculation on the “front end” has ramped up to keep pace with the reiterative work of digital repurposing on the “back end.”
Issues of intellectual property stimulate these changes. Rather than make the durable syndicatable object the company’s primary goal, spec work enterprises now obsess over the creation of potentially endless, malleable, and self-replicating IP. For clarity’s sake, we can further distinguish (especially within the same corporations) between “big” self-replicating IP (the blockbuster or high-concept), and “small” self-replicating IP (reality TV and the unexceptional online consumer interactions that go with it). As we will discover, there is now a necessary economic relationship between “big IP” and “small IP” in the transnational multimedia conglomerates. That is, diversified corporations now need vast amounts of the cheaper, reality-based small IP to pay for their expensive big-IP blockbuster and prestige cinematic needs. We need to think beyond specific tactics of content migration or repurposing to consider this broader intraconglomerate dynamic that embeds them. That is, spreadable speculation now animates and monetizes production well before—and well after—the series or episode in question. 5 This temporal spreading of pre- and post-speculation is precisely why spec work has aligned so well with transmedia production, industry–fan interactions, and viral marketing, which mirror it.
I am arguing that spec work provides the broad conditions that facilitate linkages and synergies between the malleable digital “material” and technologies of TV production, on the one hand, and current corporate management strategies aimed at developing malleable and self-replicating IP (which ideally suits corporate reformatting, franchising, branding, and transmedia), on the other hand. Before mapping out the fuller range and logic of spec work, it is worth considering something more provisional—that is, how spec work fits within the rapidly changing industrial and economic landscape. This mediascape can be usefully understood within a three-part model of craft world, brand world, and spec world.


6.3: Effective Communication
Assertiveness
Assertiveness is an honest and appropriate expression of your feelings, thoughts, wants and needs. Acting in an assertive way helps you to stand up for your rights in a respectful manner. It is a way to communicate what you believe, what you want and need, and what is important to you.
Learning to Be More AssertivePeople often associate the concept of assertiveness with standing up for your rights when you feel that someone has taken advantage of you in a negative way. However, it is also important to recognize that being more assertive can help you to communicate in a positive way in your relationships, which helps to promote mutual respect.
Assertiveness can help you:- speak up when you have a question or concern,
- say “no” when you don’t want to do something, and
- express thoughts or feelings
Aggressive, Assertive and Non-Assertive Behavior
Aggressive behavior often means standing up for yourself in ways that violate the rights of others. Aggressive behavior can be demanding, hostile, and blaming.
Non-Assertive behavior is often submissive, inhibited, passive, and self-denying.
Assertive behavior involves expressing your wants, needs, thoughts and/or feelings while respecting the rights of others.
- Not being clear about what they want and need
- Fear of displeasing others and of not being liked
- Not believing they have the right to be assertive
- Lacking the skills to effectively express themselves
“I” Statements vs. “You” Statements
“I” statements can help you focus on and be clear about your own thoughts and feelings, and what it is that you want or need. They may also involve an acknowledgement of the thoughts/feelings/goals of the other person.
The real focus in “I” statements is on the “I feel,” “I want,” or “I think” part of the statement. Identifying your thoughts, feelings, needs, and wants related to a situation will help you to avoid blaming someone else or getting caught up in the emotion of the moment.“You” statements, on the other hand, tend to place blame or criticize the other person. This typically puts the other person on the defensive, and does not encourage open communication.
For example, saying “I feel worried when you are running late to meet me for dinner and don’t call to let me know” (I statement) vs. “You are always running late, and never bother to let me know” (You statement) will likely result in two very different reactions and conversations! The first statement simply expresses how the person is feeling, whereas the second statement sets a critical and accusatory tone.
Practice
To become more skilled in communicating assertively it is important to practice. You won’t learn how to become a more assertive person just by reading one book or attending one workshop. You can practice with your friends and family. Let them know what you are doing first! Ask for help/feedback on how you’re doing.
Start Gradually In the beginning, don’t try changing your behavior in the most complex or difficult situations. Practice first in the least risky ones. Some examples:- Returning a purchased item (that you are not satisfied with) to a store for a refund
- Asking your partner/roommate/kids to help empty the dishwasher or take out the garbage
- Suggesting a movie that you would like to watch for an upcoming movie night
If you start small to enhance your chances of success, you will experience how it feels to express yourself assertively and it will be easier to move on to more challenging situations.
Keep in mind:- No one can read your mind– focus on expressing and communicating what is important to you.
- Assertiveness. Authored by : National Institutes of Health. Located at : https://www.training.nih.gov/assets/Assertiveness_Handout.pdf . License : Public Domain: No Known Copyright


8.4: A Population Proportion
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During an election year, we see articles in the newspaper that state confidence intervals in terms of proportions or percentages. For example, a poll for a particular candidate running for president might show that the candidate has 40% of the vote within three percentage points (if the sample is large enough). Often, election polls are calculated with 95% confidence, so, the pollsters would be 95% confident that the true proportion of voters who favored the candidate would be between 0.37 and 0.43: (0.40 – 0.03,0.40 + 0.03).
Investors in the stock market are interested in the true proportion of stocks that go up and down each week. Businesses that sell personal computers are interested in the proportion of households in the United States that own personal computers. Confidence intervals can be calculated for the true proportion of stocks that go up or down each week and for the true proportion of households in the United States that own personal computers.
The procedure to find the confidence interval, the sample size, the error bound, and the confidence level for a proportion is similar to that for the population mean, but the formulas are different. How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.) If \(X\) is a binomial random variable, then
\[X \sim B(n, p)\nonumber \]
where \(n\) is the number of trials and \(p\) is the probability of a success.
To form a proportion, take \(X\), the random variable for the number of successes and divide it by \(n\), the number of trials (or the sample size). The random variable \(P′ \) (read "P prime") is that proportion,
\[P' = \dfrac{X}{n}\nonumber \]
(Sometimes the random variable is denoted as \(\hat{P}\), read "P hat".)
When \(n\) is large and \(p\) is not close to zero or one, we can use the normal distribution to approximate the binomial.
\[X \sim N(np, \sqrt{npq})\nonumber \]
If we divide the random variable, the mean, and the standard deviation by \(n\), we get a normal distribution of proportions with \(P′ \), called the estimated proportion, as the random variable. (Recall that a proportion as the number of successes divided by \(n\).)
\[\dfrac{X}{n} = P' \sim N\left(\dfrac{np}{n}, \dfrac{\sqrt{npq}}{n}\right)\nonumber \]
Using algebra to simplify:
\[\dfrac{\sqrt{npq}}{n} = \sqrt{\dfrac{pq}{n}}\nonumber \]
P′ follows a normal distribution for proportions:
\[\dfrac{X}{n} = P' \sim N\left(p, \sqrt{\dfrac{pq}{n}}\right)\nonumber \]
The confidence interval has the form
\[(p′ – EBP, p′ + EBP).\nonumber \]
where
- \(EBP\) is error bound for the proportion.
- \(p′ = \dfrac{x}{n}\)
- \(p′ =\) the estimated proportion of successes ( p′ is a point estimate for p , the true proportion.)
- \(x =\) the number of successes
- \(n =\) the size of the sample
The error bound (EBP) for a proportion is
\[EBP = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber \]
where \(q\ = 1 - p'\).
This formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is \(\dfrac{\sigma}{\sqrt{n}}\). For a proportion, the appropriate standard deviation is
\[\sqrt{\dfrac{pq}{n}}.\nonumber \]
However, in the error bound formula, we use
\[\sqrt{\dfrac{p'q'}{n}}\nonumber \]
as the standard deviation, instead of
\[\sqrt{\dfrac{pq}{n}}.\nonumber \]
In the error bound formula, the sample proportions p′ and q′ are estimates of the unknown population proportions p and q . The estimated proportions \(p′\) and \(q′\) are used because \(p\) and \(q\) are not known. The sample proportions \(p′\) and \(q′\) are calculated from the data: \(p′\) is the estimated proportion of successes, and \(q′\) is the estimated proportion of failures.
The confidence interval can be used only if the number of successes \(np′\) and the number of failures \(nq′\) are both greater than five.
Normal Distribution of Proportions
For the normal distribution of proportions, the \(z\)-score formula is as follows.
If
\[P' \sim N\left(p, \sqrt{\dfrac{pq}{n}}\right)\]
then the \(z\)-score formula is
\[z = \dfrac{p'-p}{\sqrt{\dfrac{pq}{n}}} \]
Example \(\PageIndex{1}\)
Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes - they own cell phones. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of adult residents of this city who have cell phones.
Solution A
- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+ or 84 calculators (Solution B).
Let \(X =\) the number of people in the sample who have cell phones. \(X\) is binomial.
\[X \sim B(500,\dfrac{421}{500}).\nonumber \]
To calculate the confidence interval, you must find \(p′\), \(q′\), and \(EBP\).
- \(n = 500\)
- \(x =\) the number of successes \(= 421\)
\[p′ = \dfrac{x}{n} = \dfrac{421}{500} = 0.842\nonumber \]
- \(p′ = 0.842\) is the sample proportion; this is the point estimate of the population proportion.
\[q′ = 1 – p′ = 1 – 0.842 = 0.158\nonumber \]
Since \(CL = 0.95\), then
\[\alpha = 1 – CL = 1 – 0.95 = 0.05\left(\dfrac{\alpha}{2}\right) = 0.025.\nonumber \]
Then
\[z_{\dfrac{\alpha}{2}} = z_{0.025 = 1.96}\nonumber \]
Use the TI-83, 83+, or 84+ calculator command invNorm(0.975,0,1) to find \(z_{0.025}\). Remember that the area to the right of \(z_{0.025}\) is \(0.025\) and the area to the left of \(z_{0.025}\) is \(0.975\). This can also be found using appropriate commands on other calculators, using a computer, or using a Standard Normal probability table.
\[EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{(0.842)(0.158)}{500}} = 0.032\nonumber \]
\[p' – EBP = 0.842 – 0.032 = 0.81\nonumber \]
\[p′ + EBP = 0.842 + 0.032 = 0.874\nonumber \]
The confidence interval for the true binomial population proportion is \((p′ – EBP, p′ +EBP) = (0.810, 0.874)\).
Interpretation
We estimate with 95% confidence that between 81% and 87.4% of all adult residents of this city have cell phones.
Explanation of 95% Confidence Level
Ninety-five percent of the confidence intervals constructed in this way would contain the true value for the population proportion of all adult residents of this city who have cell phones.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to xx and enter 421.
Arrow down to nn and enter 500.
Arrow down to
C-Level
and enter .95.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.81003, 0.87397).
Exercise \(\PageIndex{1}\)
Suppose 250 randomly selected people are surveyed to determine if they own a tablet. Of the 250 surveyed, 98 reported owning a tablet. Using a 95% confidence level, compute a confidence interval estimate for the true proportion of people who own tablets.
- Answer
-
(0.3315, 0.4525)
Example \(\PageIndex{2}\)
For a class project, a political science student at a large university wants to estimate the percent of students who are registered voters. He surveys 500 students and finds that 300 are registered voters. Compute a 90% confidence interval for the true percent of students who are registered voters, and interpret the confidence interval.
Answer
- The first solution is step-by-step (Solution A).
- The second solution uses a function of the TI-83, 83+, or 84 calculators (Solution B).
Solution A
- \(x = 300\) and
- \(n = 500\)
\[p' = \dfrac{x}{n} = \dfrac{300}{500} = 0.600\nonumber \]
\[q′ = 1 − p′ = 1 − 0.600 = 0.400\nonumber \]
Since \(CL = 0.90\), then
\[\alpha = 1 – CL = 1 – 0.90 = 0.10\left(\dfrac{\alpha}{2}\right) = 0.05\]
\[z_{\dfrac{\alpha}{2}} = z_{0.05} = 1.645\nonumber \]
Use the TI-83, 83+, or 84+ calculator command invNorm(0.95,0,1) to find \(z_{0.05}\). Remember that the area to the right of \(z_{0.05}\) is 0.05 and the area to the left of \(z_{0.05}\) is 0.95. This can also be found using appropriate commands on other calculators, using a computer, or using a standard normal probability table.
\[EBP = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.645)\sqrt{\dfrac{(0.60)(0.40)}{500}} = 0.036\nonumber \]
\[p′ – EBP = 0.60 − 0.036 = 0.564\nonumber \]
\[p′ + EBP = 0.60 + 0.036 = 0.636\nonumber \]
The confidence interval for the true binomial population proportion is \((p′ – EBP, p′ +EBP) = (0.564,0.636)\).
Interpretation
- We estimate with 90% confidence that the true percent of all students that are registered voters is between 56.4% and 63.6%.
- Alternate Wording: We estimate with 90% confidence that between 56.4% and 63.6% of ALL students are registered voters.
Explanation of 90% Confidence Level
Ninety percent of all confidence intervals constructed in this way contain the true value for the population percent of students that are registered voters.
Solution B
Press
STAT
and arrow over to
TESTS
.
Arrow down to
A:1-PropZint
. Press
ENTER
.
Arrow down to xx and enter 300.
Arrow down to nn and enter 500.
Arrow down to
C-Level
and enter 0.90.
Arrow down to
Calculate
and press
ENTER
.
The confidence interval is (0.564, 0.636).
Exercise \(\PageIndex{2}\)
A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.
- Compute a 90% confidence interval for the true percent of students who are against the new legislation, and interpret the confidence interval.
- In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.
- Answer a
-
(0.7731, 0.8269); We estimate with 90% confidence that the true percent of all students in the district who are against the new legislation is between 77.31% and 82.69%.
- Answer b
-
Sixty-eight percent (68%) of students own an iPod and a smart phone.
\[p′ = 0.68\nonumber \]
\[q′ = 1–p′ = 1 – 0.68 = 0.32\nonumber \]
Since \(CL = 0.97\), we know
\[\alpha = 1 – 0.97 = 0.03\nonumber \]
and
\[\dfrac{\alpha}{2} = 0.015.\nonumber \]
The area to the left of \(z_{0.05}\) is 0.015, and the area to the right of \(z_{0.05}\) is 1 – 0.015 = 0.985.
Using the TI 83, 83+, or 84+ calculator function InvNorm(0.985,0,1),
\[z_{0.05} = 2.17\nonumber \]
\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = 2.17\sqrt{\dfrac{0.68(0.32)}{300}} \approx 0.0269\nonumber \]
\[p′ – EPB = 0.68 – 0.0269 = 0.6531\nonumber \]
\[p′ + EPB = 0.68 + 0.0269 = 0.7069\nonumber \]
We are 97% confident that the true proportion of all students who own an iPod and a smart phone is between 0.6531 and 0.7069.
Calculator
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to x and enter 300*0.68.
Arrow down to n and enter 300.
Arrow down to C-Level and enter 0.97.
Arrow down to Calculate and press ENTER.The confidence interval is (0.6531, 0.7069).
"Plus Four" Confidence Interval for \(p\)
There is a certain amount of error introduced into the process of calculating a confidence interval for a proportion. Because we do not know the true proportion for the population, we are forced to use point estimates to calculate the appropriate standard deviation of the sampling distribution. Studies have shown that the resulting estimation of the standard deviation can be flawed.
Fortunately, there is a simple adjustment that allows us to produce more accurate confidence intervals. We simply pretend that we have four additional observations. Two of these observations are successes and two are failures. The new sample size, then, is \(n + 4\), and the new count of successes is \(x + 2\). Computer studies have demonstrated the effectiveness of this method. It should be used when the confidence level desired is at least 90% and the sample size is at least ten.
Example \(\PageIndex{3}\)
A random sample of 25 statistics students was asked: “Have you smoked a cigarette in the past week?” Six students reported smoking within the past week. Use the “plus-four” method to find a 95% confidence interval for the true proportion of statistics students who smoke.
Solution A
Six students out of 25 reported smoking within the past week, so \(x = 6\) and \(n = 25\). Because we are using the “plus-four” method, we will use \(x = 6 + 2 = 8\) and \(n = 25 + 4 = 29\).
\[p' = \dfrac{x}{n} = \dfrac{8}{29} \approx 0.276\nonumber \]
\[q′ = 1 – p′ = 1 – 0.276 = 0.724\nonumber \]
Since \(CL = 0.95\), we know \(\alpha = 1 – 0.95 = 0.05\) and \(\dfrac{\alpha}{2} = 0.025\).
\[z_{0.025} = 1.96\nonumber \]
\(EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (1.96)\sqrt{\dfrac{0.276(0.724)}{29}} \approx 0.163\)
\[p′ – EPB = 0.276 – 0.163 = 0.113\nonumber \]
\[p′ + EPB = 0.276 + 0.163 = 0.439\nonumber \]
We are 95% confident that the true proportion of all statistics students who smoke cigarettes is between 0.113 and 0.439.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
REMINDER
Remember that the plus-four method assume an additional four trials: two successes and two failures. You do not need to change the process for calculating the confidence interval; simply update the values of x and n to reflect these additional trials.
Arrow down to \(x\) and enter eight.
Arrow down to \(n\) and enter 29.
Arrow down to C-Level and enter 0.95.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.113, 0.439).
Exercise \(\PageIndex{3}\)
Out of a random sample of 65 freshmen at State University, 31 students have declared a major. Use the “plus-four” method to find a 96% confidence interval for the true proportion of freshmen at State University who have declared a major.
Solution A
Using “plus four,” we have \(x = 31 + 2 = 33\) and \(n = 65 + 4 = 69\).
\[p′ = 3369 \approx 0.478\nonumber \]
\[q′ = 1 – p′ = 1 – 0.478 = 0.522\nonumber \]
Since \(CL = 0.96\), we know \(\alpha = 1 – 0.96 = 0.04\) and \(\dfrac{\alpha}{2} = 0.02\).
\[z_{0.02} = 2.054\nonumber \]
\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\sqrt{\dfrac{p'q'}{n}} = (2.054)\left(\sqrt{\dfrac{(0.478)(0.522)}{69}}\right) - 0.124\nonumber \]
\[p′ – EPB = 0.478 – 0.124 = 0.354\nonumber \]
\[p′ + EPB = 0.478 + 0.124 = 0.602\nonumber \]
We are 96% confident that between 35.4% and 60.2% of all freshmen at State U have declared a major.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to \(x\) and enter 33.
Arrow down to \(n\) and enter 69.
Arrow down to C-Level and enter 0.96.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.355, 0.602).
Example \(\PageIndex{4}\)
The Berkman Center for Internet & Society at Harvard recently conducted a study analyzing the privacy management habits of teen internet users. In a group of 50 teens, 13 reported having more than 500 friends on Facebook. Use the “plus four” method to find a 90% confidence interval for the true proportion of teens who would report having more than 500 Facebook friends.
Solution A
Using “plus-four,” we have \(x = 13 + 2 = 15\) and \(n = 50 + 4 = 54\).
\[p′ = 1554 \approx 0.278\nonumber \]
\[q′ = 1 – p′ = 1 − 0.241 = 0.722\nonumber \]
Since \(CL = 0.90\), we know \(\alpha = 1 – 0.90 = 0.10\) and \(\dfrac{\alpha}{2} = 0.05\).
\[z_{0.05} = 1.645\nonumber \]
\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.278)(0.722)}{54}}\right) \approx 0.100\nonumber \]
\[p′ – EPB = 0.278 – 0.100 = 0.178\nonumber \]
\[p′ + EPB = 0.278 + 0.100 = 0.378\nonumber \]
We are 90% confident that between 17.8% and 37.8% of all teens would report having more than 500 friends on Facebook.
Solution B
Press STAT and arrow over to TESTS.
Arrow down to A:1-PropZint. Press ENTER.
Arrow down to \(x\) and enter 15.
Arrow down to \(n\) and enter 54.
Arrow down to C-Level and enter 0.90.
Arrow down to Calculate and press ENTER.
The confidence interval is (0.178, 0.378).
Exercise \(\PageIndex{4}\)
The Berkman Center Study referenced in Example talked to teens in smaller focus groups, but also interviewed additional teens over the phone. When the study was complete, 588 teens had answered the question about their Facebook friends with 159 saying that they have more than 500 friends. Use the “plus-four” method to find a 90% confidence interval for the true proportion of teens that would report having more than 500 Facebook friends based on this larger sample. Compare the results to those in Example.
Answer
Solution A
Using “plus-four,” we have \(x = 159 + 2 = 161\) and \(n = 588 + 4 = 592\).
\[p′ = 161592 \approx 0.272\nonumber \]
\[q′ = 1 – p′ = 1 – 0.272 = 0.728\nonumber \]
Since CL = 0.90, we know \(\alpha = 1 – 0.90 = 0.10\) and \(\dfrac{\alpha}{2} = 0.05\)
\[EPB = \left(z_{\dfrac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right) = (1.645)\left(\sqrt{\dfrac{(0.272)(0.728)}{592}}\right) \approx 0.030\nonumber \]
\[p′ – EPB = 0.272 – 0.030 = 0.242\nonumber \]
\[p′ + EPB = 0.272 + 0.030 = 0.302\nonumber \]
We are 90% confident that between 24.2% and 30.2% of all teens would report having more than 500 friends on Facebook.
Solution B
- Press STAT and arrow over to TESTS.
- Arrow down to A:1-PropZint. Press ENTER.
- Arrow down to \(x\) and enter 161.
- Arrow down to \(n\) and enter 592.
- Arrow down to C-Level and enter 0.90.
- Arrow down to Calculate and press ENTER.
- The confidence interval is (0.242, 0.302).
Conclusion: The confidence interval for the larger sample is narrower than the interval from Example. Larger samples will always yield more precise confidence intervals than smaller samples. The “plus four” method has a greater impact on the smaller sample. It shifts the point estimate from 0.26 (13/50) to 0.278 (15/54). It has a smaller impact on the EPB , changing it from 0.102 to 0.100. In the larger sample, the point estimate undergoes a smaller shift: from 0.270 (159/588) to 0.272 (161/592). It is easy to see that the plus-four method has the greatest impact on smaller samples.
Calculating the Sample Size \(n\)
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size. The error bound formula for a population proportion is
\[EBP = \left(z_{\frac{\alpha}{2}}\right)\left(\sqrt{\dfrac{p'q'}{n}}\right)\nonumber \]
Solving for \(n\) gives you an equation for the sample size.
\[n = \dfrac{\left(z_{\frac{\alpha}{2}}\right)^{2}(p'q')}{EBP^{2}}\nonumber \]
Example \(\PageIndex{5}\)
Suppose a mobile phone company wants to determine the current percentage of customers aged 50+ who use text messaging on their cell phones. How many customers aged 50+ should the company survey in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of customers aged 50+ who use text messaging on their cell phones.
Answer
From the problem, we know that \(\bf{EBP = 0.03}\) (3%=0.03) and \(z_{\dfrac{\alpha}{2}} z_{0.05} = 1.645\) because the confidence level is 90%.
However, in order to find \(n\), we need to know the estimated (sample) proportion \(p′\). Remember that \(q′ = 1 – p′\). But, we do not know \(p′\) yet. Since we multiply \(p′\) and \(q′\) together, we make them both equal to 0.5 because \(p′q′ = (0.5)(0.5) = 0.25\) results in the largest possible product. (Try other products: \((0.6)(0.4) = 0.24\); \((0.3)(0.7) = 0.21\); \((0.2)(0.8) = 0.16\) and so on). The largest possible product gives us the largest \(n\). This gives us a large enough sample so that we can be 90% confident that we are within three percentage points of the true population proportion. To calculate the sample size \(n\), use the formula and make the substitutions.
\[n = \dfrac{z^{2}p'q'}{EBP^{2}}\nonumber \]
gives
\[n = \dfrac{1.645^{2}(0.5)(0.5)}{0.03^{2}} = 751.7\nonumber \]
Round the answer to the next higher value. The sample size should be 752 cell phone customers aged 50+ in order to be 90% confident that the estimated (sample) proportion is within three percentage points of the true population proportion of all customers aged 50+ who use text messaging on their cell phones.
Exercise \(\PageIndex{5}\)
Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 90% confident that the estimated proportion is within five percentage points of the true population proportion of customers who click on ads on their smartphones?
- Answer
-
271 customers should be surveyed. Check the Real Estate section in your local
Glossary
- Binomial Distribution
- a discrete random variable (RV) which arises from Bernoulli trials; there are a fixed number, \(n\), of independent trials. “Independent” means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV \(X\) is defined as the number of successes in \(n\) trials. The notation is: \(X \sim B(\mathbf{n},\mathbf{p})\). The mean is \(\mu = np\) and the standard deviation is \(\sigma = \sqrt{npq}\). The probability of exactly \(x\) successes in \(n\) trials is \(P(X = x = \left(\binom{n}{x}\right))p^{x}q^{n-x}\).
- Error Bound for a Population Proportion (\(EBP\))
- the margin of error; depends on the confidence level, the sample size, and the estimated (from the sample) proportion of successes.